Particular Point Space is Non-Meager/Proof 3

From ProofWiki
Jump to navigation Jump to search


Let $T = \left({S, \tau_p}\right)$ be a particular point space.

Then $T$ is non-meager.


By definition of particular point space, any subset of $S$ which contains $p$ is open in $T$.

Aiming for a contradiction, suppose $T$ is meager.

By definition, $T$ is meager if and only if it is a countable union of subsets of $S$ which are nowhere dense in $T$.

At least one such nowhere dense subset $U$ of $S$ must contain $p$.

By definition, $U$ is nowhere dense in $T$ if and only if:

$U^-$ contains no open set of $T$ which is non-empty

where $U^-$ denotes the closure of $U$.

By definition of particular point space, $U$ is open in $T$.

By Closure of Open Set of Particular Point Space, $U^- = S$.

But $S$ is itself open in $T$ and non-empty, and so $U$ is not nowhere dense.

From this contradiction it follows that $T$ is non-meager